$\newcommand{\tb}{2_{\!\bot}}\newcommand{\tbO}{\tb^{\,\omega}}$Let $2 = \{0,1\}$ and $\tb = \{0,\bot,1\}$.
Scott continuity is important for defining models of lambda calculus, a formalism for defining Turing-computable functions. That is because every Turing/lambda-computable function is Scott-continuous, and there are domains such that $$D \cong [D \to D]\,,$$ where $[D \to D]$ is the set of Scott-continuous functions on $D$.
Infinite time Turing machines (ITTMs) are more expressive than Turing machines, and are capable of computing functions that are not Scott-continuous. Take the function $z\colon 2^\omega \to 2$ that returns 0 iff its argument is a tape of zeros. (An ITTM may cycle moving left, return 1 if it finds 1 on the tape, or return 0 if it finds itself in the limit state. However, $z$ is not Scott-continuous, see below.)
Question: Is there some other continuity $Y$ that would similarly be to ITTM-computable functions what Scott continuity is to Turing-computable functions?
It would have to be the case that all ITTM-computable functions are $Y$-continuous. Presumably, there would also be domains isomorphic to their $Y$-continuous function spaces, containing all ITTM-computable functions. (I don't know what other, if any, requirements I should have.)
Definitions:
Let $<$ be a partial order on $\tb$ such that only $\bot < 0$ and $\bot < 1$ hold. Let $\tbO$ be ordered pointwise.
A pointed directed-complete partial order (CPO) is a partially ordered set with a least element, such that every directed set has a supremum. $\tbO$ and $\tb$ are CPOs.
Let $D$ and $E$ be CPOs. A function $f\colon D \to E$ is Scott continuous iff for every directed subset $P$ of $D$, its image has a supremum $\sqcup f(P)$ and it equals $f(\sqcup P)$.
I believe it is reasonable to extend the definition of Scott continuity to functions $g\colon 2^\omega \to 2$ as follows, but if I'm wrong here, please correct me: $g$ is continuous iff the following function is: \begin{align} g'\colon \tbO &\to \tb \\ x &\mapsto \sqcap g(\{ y \in 2^\omega \mid x \leq y \}) \end{align}
(I believe this corresponds with how continuity of $g$ is defined in this answer using the product topology on $2^\omega$, and has the intuitive explanation that we are treating elements of $\tb^\omega$ as sequences of finite pieces of information about some element of $2^\omega$, each piece telling us the value of the corresponding digit (if it's 0 or 1) or that the digit is unknown (if it's $\bot$). If this is unreasonable (or incorrect), then unfortunately I don't know how one might extend the definition of Scott continuity to functions from $2^\omega$ to $2$; hopefully my question can still be salvaged.)
With this definition, we might show why $z$ from above is not Scott-continuous: for every $n \in \mathbb{N}$, let $Z_n = 0^n \cdot \bot^\omega$, where $\cdot$ is concatenation. For every $n$, $z'(Z_n) = \bot$, so $\sqcup_n z'(Z_n) = \bot$, however $z'(\sqcup_n Z_n) = z'(0^\omega) = 0$.